$\sum_{n=1}^\infty \frac{a_1+a_2+\cdots+a_n}{n}$ converges/divereges?

Let $$a_n\geq 0$$. $$\sum_{n=1}^\infty \frac{a_1+a_2+\cdots+a_n}{n}$$ converges/divereges? Take $$a_n=1/n$$ for example, then the proposed series looks like $$\sum\frac{\ln n}{n}$$, which is divergent. What about the general case?

• it cud also converge obviously; e.g. $a_n = 0$ for each $n$ – mathworker21 Jun 9 '19 at 12:39
• Or even worse: $\frac{1+2+3+\cdots+n}{n}$ clearly diverges. You need an extra condition such as $\lim_{n \to \infty} a_n = 0$ or that $a_n$ is decreasing. – Toby Mak Jun 9 '19 at 12:40
• If $a_1 > 0$ then the sum is at least $\sum_{n = 1}^{\infty}\tfrac{a_1}{n}$ which diverges. Similarly if any of the $a_n$'s are positive, the sum diverges. – JimmyK4542 Jun 9 '19 at 12:41
• looks like cauchy's first theorem on limit. see here math.stackexchange.com/questions/1930373/… – aud098 Jan 24 '20 at 16:33

Convergence holds only when $$a_n=0$$ for all $$n$$. $$\sum\limits_{n=1}^{\infty} \frac 1 n \sum\limits_{i=1}^{n} a_i=\sum\limits_{i=1}^{\infty}a_i \sum\limits_{n=i}^{\infty} \frac 1 n=\infty$$ unless $$a_i=0$$ for all $$i$$.
• [ I have used the fact that $\sum\limits_{n=i}^{\infty} \frac 1 n = \infty$ for every $i$]. – Kavi Rama Murthy Jun 9 '19 at 12:48